Chapter 3 Polynomial Functions
Section 3.4 Zeros of Polynomial Functions
Course/Section
Lesson Number
Date
Section Objectives: Students will know how to determine the number of
rational and real zeros of polynomial functions, and how to
find the zeros.
I. The Fundamental Theorem of Algebra (p. 293)
Pace: 5 minutes
· State the Fundamental Theorem of Algebra.
If f(x) is a polynomial function of degree n, where n > 0, then f has at
least one zero in the complex number system.
· State the Linear Factorization Theorem.
If f(x) is a polynomial function of degree n, where n > 0, then f has
precisely n linear factors
f(x) = an(x – c1) (x – c2)L (x – cn)
where c1, c2, K, cn are complex zeros.
Tip: It should be pointed out that the zeros in the above theorem may not be
distinct.
II. The Rational Zero Test (pp. 294-296)
Pace: 15 minutes
· Ask the students how they would solve x3 + 6x – 7 = 0. Then ask them how
they would solve the same equation if they knew that, if there were any, the
rational zeros would have to be in the list ±1, ±2, ±3, ±6. Now state the
Rational Zero Test.
If the polynomial f(x) = anxn + an-1xn-1 + L + a1 + a0 has integer
coefficients with an ¹ 0 and a0 ¹ 0, then any rational zero of f will be of
the form p/q, where p is a factor of a0 and q is a factor of an.
Example 1. Find the zeros of x3 - 7x – 6 = 0.
p: ±1, ±2, ±3, ±6
q: ±1
p/q: ±1, ±2, ±3, ±6
Use synthetic division to find a number from the list that is a solution.
-1 1 0 -7 -6
-1 1 6
1 -1 -6 0
We now have (x + 1)(x2 - x - 6) = 0. x – 1 = 0 Þ x = 1.
x2 - x - 6 = (x – 3)(x + 2) = 0 Þ x = -2 or x = 3.
The zeros are x = 1, x = -2, and x = 3.
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3.4-1
Example 2. Find all real zeros of 3x3 – 20x2 + 23x + 10.
p: ±1, ±2, ±5, ±10
q: ±1, ±3
p/q: ±1, ±2, ±5, ±10, ±1/3, ±2/3, ±5/3, ±10/3
2 3 -20 23 10
6 -28 -10
3 -14 -5 0
One zero is 2. Two more come from solving
3 x 2 -14 x - 5 = 0
(3 x +1)(x - 5) = 0
x-5=0Þ x =5
3x + 1 = 0 Þ x = -
1
3
III. Conjugate Pairs (p. 297)
Pace: 5 minutes
· Note that in Example 1(c) and (d) of the text, the two complex zeros were
conjugates. State that if f is a polynomial function with real coefficients,
then whenever a + bi is a zero of f, a – bi is also a zero of f.
Example 3. Find a fourth-degree polynomial function with real
coefficients that has 0, 1, and i as zeros.
Since i is a zero, -i is also a zero.
f(x) = x(x – 1)(x – i)(x + i) = x4 – x3 + x2 – x
IV. Factoring a Polynomial (pp. 297-299)
Pace: 5 minutes
· State that the Linear Factorization Theorem, together with the above
statement regarding complex zeros and conjugate pairs, leads to the
following statement regarding factoring a polynomial over the reals.
Every polynomial of degree n > 0 with real coefficients can be written
as the product of linear and quadratic factors with real coefficients,
where the quadratic factors have no real zeros.
Example 4. Find all zeros of f(x) = x4 – 4x3 + 12x2 + 4x – 13, given
that 2 + 3i is a zero.
Because 2 + 3i is a zero, 2 – 3i is also a zero. This means that x2 – 4x +
13 is a factor of f(x).
2
x -1
x 2 - 4x + 13 x 4 - 4x 3 + 12 x 2 + 4 x - 13
)
4
3
x - 4x + 13x
2
- x 2 + 4 x - 13
2
-x + 4 x -13
0
All the zeros of f are –1, 1, 2 + 3i, 2 – 3i.
3.4-2
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V. Other Tests for Zeros of Polynomials (pp. 300-302)
Pace: 10 minutes
·
There are a couple of ways of dealing with a very large list generated by the
Rational Zero Test. The first of these is Descartes’s Rule of Signs, which
states:
Let f(x) = anxn + an-1xn-1 + L + a1x + a0 be a polynomial with real
coefficients and a0 ¹ 0.
1. The number of positive real zeros of f is either equal to the
number of variations in sign of f(x) or less than that number by an
even integer.
2. The number of negative real zeros of f is either equal to the
number of variations in sign of f(-x) or less than that number by
an even integer.
Two notes about Descartes’s Rule of Signs:
1. A variation in sign means that two consecutive coefficients have
opposite signs.
2. When we count the zeros, we must count their multiplicities.
Example 5. Describe the possible real zeros of
f(x) = 7x3 + 3x2 – 5x + 9.
f(x) has two variations in sign; therefore, there are either two or no
positive real zeros.
f(-x) = -7x3 + 3x2 + 5x + 9.
f(-x) has one variation in sign; therefore, there is exactly one negative
real zero.
·
·
The second way of dealing with a very large list generated by the Rational
Zero Test is the Upper and Lower Bound Rules. Before you state this
rule, discuss what upper and lower bounds are. A real number b is an
upper bound for the real zeros of f if there are no zeros of f greater than b.
A real number b is a lower bound for the real zeros of f if there are no zeros
of f less than b.
Upper and Lower Bound Rules
Let f(x) be a polynomial function with real coefficients and a positive
leading coefficient. Suppose f(x) is divided by x – c using synthetic
division.
1. If c > 0 and each number in the last row is either positive or zero, then c
is an upper bound for the real zeros of f.
2. If c < 0 and the numbers in the last row are alternately positive and
negative (zero entries count as either positive or negative), then c is a
lower bound for the real zeros of f.
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Example 6. Find all real zeros of f(x) = x4 - 3x3 + x - 3.
p: ±1, ±3
q: ±1
p/q: ±1, ±3
f(x) has three variations in sign; therefore f(x) has 3 or 1 positive real
zeros.
f(-x) has one variation in sign; therefore f(x) has exactly 1 negative real
zero. We will start by trying to find it.
-1 1 -3 0 1 -3
-1 4 -4 3
1 -4 4 -3 0
Now we look for the positive zero. Testing the value 1 does not work,
so 3 must be a zero.
3 1 -4 4 -3
3 -3 3
1 -1 1 0
Now we solve x2 – x + 1 = 0 by using the quadratic formula.
x=
- (-1) ±
=
1 ± -3
2
=
1
3
±
i
2
2
(-1)2 - 4(1)(1)
2(1)
Therefore, all four zeros of f are –1, 3, and
3.4-4
1
3
±
i.
2
2
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